Solution (source code)

= Solution

At $r=s=0$, the linear growth rate of a <Fourier mode> $e^{ikx}$ is $-(1-k^2)^2$. Its critical <kernel of a linear map> has $k=\pm1$. Write
$$
w_1=A(X,T)e^{ix}+\overline{A(X,T)}e^{-ix}.
$$
For a compatible periodic carrier, the slowly varying envelope is periodic on $0\le X\le\varepsilon^2L$. Small carrier detuning can alternatively be included in the envelope <wave phase>.

With $w=O(\varepsilon)$, $r=\varepsilon^4\mu$ and $s=\varepsilon^2\hat s$, all three terms $rw$, $sw^3$, $w^5$ first appear at $O(\varepsilon^5)$. Balancing the time <derivative> thus requires $T=\varepsilon^4t$. Since the linear growth rate near $k=1$ is $-4(k-1)^2+O((k-1)^3)$, spatial detuning must be $O(\varepsilon^2)$ to contribute at the same order. Consequently
$$
\boxed{X=\varepsilon^2x,\qquad T=\varepsilon^4t.}
$$
Treating $x,X$ independently, let $D=\partial_x$ on the fast carrier. Expansion of the linear operator gives
$$
(1+(D+\varepsilon^2\partial_X)^2)^2
=(1+D^2)^2+4\varepsilon^2(1+D^2)D\partial_X
+\varepsilon^4(2+6D^2)\partial_X^2+O(\varepsilon^6).
$$
The $O(\varepsilon^3)$ term annihilates $w_1$, since $(1+D^2)w_1=0$. At orders two through four there is no nonresonant forcing, and <kernel of a linear map> corrections can be absorbed into the definition of $A$, allowing $w_2=w_3=w_4=0$. At order five the equation for $w_5$ is
$$
(1+D^2)^2w_5=\mu w_1+\hat s w_1^3-w_1^5+4w_{1,XX}-w_{1,T}.
$$
The fast operator is a <self-adjoint operator> on periodic functions. Its <Fredholm alternative> requires the right-hand side to have zero <inner product> with each <kernel of a linear map> mode $e^{\pm ix}$, equivalently to have zero resonant <Fourier coefficients>. The coefficient of $e^{ix}$ in $w_1^3$ is $3A^2\overline A$, and in $w_1^5$ it is $\binom53A^3\overline A^2=10A|A|^4$. Therefore the <solvability condition> is
$$
\boxed{A_T=\mu A+3\hat s A|A|^2-10A|A|^4+4A_{XX}.}
$$
The conjugate condition supplies the same real equation for the negative mode. The remaining noncritical harmonics determine $w_5$ without obstruction. This is the <cubic-quintic Swift–Hohenberg amplitude reduction>; it retains quintic saturation because the small positive cubic coefficient makes the onset mildly subcritical.